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SSC संयुक्त स्नातक स्तरीय टीयर-1 प्रैक्टिस सेट 1

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Question : 71 of 100

Marks: +1, -0

यदि

\sin^{8}\theta+\cos^{8}\theta-1=0

\cos^{2}\theta\sin^{2}\theta

का मान क्या है (यदि

\theta\neq0

\frac{\pi}{2}) ?

$-1$
0
1
2

Solution:

\sin^{8}\theta+\cos^{8}\theta-1=0

\Rightarrow(\sin^{4}\theta)^{2}+(\cos^{4}\theta)^{2}+2\sin^{4}\theta

\cdot\cos^{4}\theta-2\sin^{4}\theta\cdot\cos^{4}\theta-1=0

\Rightarrow(\sin^{4}\theta+\cos^{4}\theta)^{2}-2\sin^{4}\theta

\cdot\cos^{4}\theta-1=0

\Rightarrow[(\sin^{2}\theta)^{2}+(\cos^{2}\theta)^{2}+2\sin^{2}\theta

\cos^{2}\theta-2\sin^{2}\theta\cdot\cos^{2}\theta]^{2}

-2\sin^{4}\theta\cdot\cos^{4}\theta-1=0

\Rightarrow[(\sin^{2}\theta+\cos^{2}\theta)^{2}-2\sin^{2}\theta

\cdot\cos^{2}\theta]^{2}-2\sin^{4}\theta\cdot\cos^{4}\theta-1=0

\because\sin^{2}\theta+\cos^{2}\theta=1

\therefore[1-2\sin^{2}\theta\cdot\cos^{2}\theta]^{2}

-2\sin^{4}\theta\cdot\cos^{4}\theta-1=0

\Rightarrow[1+4\sin^{4}\theta\cdot\cos^{4}\theta-2

\times 2\sin^{2}\theta\cdot\cos^{2}\theta-2\sin^{4}\theta

\cdot\cos^{4}\theta-1=0

\Rightarrow 2\sin^{4}\theta\cdot\cos^{4}\theta-4\sin^{2}\theta

\cdot\cos^{2}\theta=0

\Rightarrow2\sin^{2}\theta\cdot\cos^{2}\theta

(\sin^{2}\theta\cdot\cos^{2}\theta-2)=0

\Rightarrow2\sin^{2}\theta\cdot\cos^{2}\theta=0

\therefore\sin^{2}\theta\cdot\cos^{2}\theta=0

(असत्य)

\because\theta\neq0

अथवा

90^{\circ}

\Rightarrow\sin^{2}\theta\cdot\cos^{2}\theta-2=0

\Rightarrow\sin^{2}\theta\cdot\cos^{2}\theta=2

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